Optimal. Leaf size=244 \[ \frac {(A+i B) \text {ArcTan}\left (\frac {\sqrt {i a-b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{(i a-b)^{5/2} d}+\frac {(A-i B) \tanh ^{-1}\left (\frac {\sqrt {i a+b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{(i a+b)^{5/2} d}+\frac {2 a (A b-a B) \sqrt {\tan (c+d x)}}{3 b \left (a^2+b^2\right ) d (a+b \tan (c+d x))^{3/2}}+\frac {2 \left (2 a^2 A b-4 A b^3+a^3 B+7 a b^2 B\right ) \sqrt {\tan (c+d x)}}{3 b \left (a^2+b^2\right )^2 d \sqrt {a+b \tan (c+d x)}} \]
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Rubi [A]
time = 0.65, antiderivative size = 244, normalized size of antiderivative = 1.00, number of steps
used = 9, number of rules used = 7, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {3686, 3730,
3697, 3696, 95, 209, 212} \begin {gather*} \frac {2 a (A b-a B) \sqrt {\tan (c+d x)}}{3 b d \left (a^2+b^2\right ) (a+b \tan (c+d x))^{3/2}}+\frac {2 \left (a^3 B+2 a^2 A b+7 a b^2 B-4 A b^3\right ) \sqrt {\tan (c+d x)}}{3 b d \left (a^2+b^2\right )^2 \sqrt {a+b \tan (c+d x)}}+\frac {(A+i B) \text {ArcTan}\left (\frac {\sqrt {-b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d (-b+i a)^{5/2}}+\frac {(A-i B) \tanh ^{-1}\left (\frac {\sqrt {b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d (b+i a)^{5/2}} \end {gather*}
Antiderivative was successfully verified.
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Rule 95
Rule 209
Rule 212
Rule 3686
Rule 3696
Rule 3697
Rule 3730
Rubi steps
\begin {align*} \int \frac {\tan ^{\frac {3}{2}}(c+d x) (A+B \tan (c+d x))}{(a+b \tan (c+d x))^{5/2}} \, dx &=\frac {2 a (A b-a B) \sqrt {\tan (c+d x)}}{3 b \left (a^2+b^2\right ) d (a+b \tan (c+d x))^{3/2}}+\frac {2 \int \frac {-\frac {1}{2} a (A b-a B)+\frac {3}{2} b (A b-a B) \tan (c+d x)+\frac {1}{2} \left (2 a A b+a^2 B+3 b^2 B\right ) \tan ^2(c+d x)}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}} \, dx}{3 b \left (a^2+b^2\right )}\\ &=\frac {2 a (A b-a B) \sqrt {\tan (c+d x)}}{3 b \left (a^2+b^2\right ) d (a+b \tan (c+d x))^{3/2}}+\frac {2 \left (2 a^2 A b-4 A b^3+a^3 B+7 a b^2 B\right ) \sqrt {\tan (c+d x)}}{3 b \left (a^2+b^2\right )^2 d \sqrt {a+b \tan (c+d x)}}+\frac {4 \int \frac {-\frac {3}{4} a b \left (a^2 A-A b^2+2 a b B\right )+\frac {3}{4} a b \left (2 a A b-a^2 B+b^2 B\right ) \tan (c+d x)}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}} \, dx}{3 a b \left (a^2+b^2\right )^2}\\ &=\frac {2 a (A b-a B) \sqrt {\tan (c+d x)}}{3 b \left (a^2+b^2\right ) d (a+b \tan (c+d x))^{3/2}}+\frac {2 \left (2 a^2 A b-4 A b^3+a^3 B+7 a b^2 B\right ) \sqrt {\tan (c+d x)}}{3 b \left (a^2+b^2\right )^2 d \sqrt {a+b \tan (c+d x)}}-\frac {(A-i B) \int \frac {1+i \tan (c+d x)}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}} \, dx}{2 (a-i b)^2}-\frac {(A+i B) \int \frac {1-i \tan (c+d x)}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}} \, dx}{2 (a+i b)^2}\\ &=\frac {2 a (A b-a B) \sqrt {\tan (c+d x)}}{3 b \left (a^2+b^2\right ) d (a+b \tan (c+d x))^{3/2}}+\frac {2 \left (2 a^2 A b-4 A b^3+a^3 B+7 a b^2 B\right ) \sqrt {\tan (c+d x)}}{3 b \left (a^2+b^2\right )^2 d \sqrt {a+b \tan (c+d x)}}-\frac {(A-i B) \text {Subst}\left (\int \frac {1}{(1-i x) \sqrt {x} \sqrt {a+b x}} \, dx,x,\tan (c+d x)\right )}{2 (a-i b)^2 d}-\frac {(A+i B) \text {Subst}\left (\int \frac {1}{(1+i x) \sqrt {x} \sqrt {a+b x}} \, dx,x,\tan (c+d x)\right )}{2 (a+i b)^2 d}\\ &=\frac {2 a (A b-a B) \sqrt {\tan (c+d x)}}{3 b \left (a^2+b^2\right ) d (a+b \tan (c+d x))^{3/2}}+\frac {2 \left (2 a^2 A b-4 A b^3+a^3 B+7 a b^2 B\right ) \sqrt {\tan (c+d x)}}{3 b \left (a^2+b^2\right )^2 d \sqrt {a+b \tan (c+d x)}}-\frac {(A-i B) \text {Subst}\left (\int \frac {1}{1-(i a+b) x^2} \, dx,x,\frac {\sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{(a-i b)^2 d}-\frac {(A+i B) \text {Subst}\left (\int \frac {1}{1-(-i a+b) x^2} \, dx,x,\frac {\sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{(a+i b)^2 d}\\ &=\frac {(A+i B) \tan ^{-1}\left (\frac {\sqrt {i a-b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{(i a-b)^{5/2} d}+\frac {(A-i B) \tanh ^{-1}\left (\frac {\sqrt {i a+b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{(i a+b)^{5/2} d}+\frac {2 a (A b-a B) \sqrt {\tan (c+d x)}}{3 b \left (a^2+b^2\right ) d (a+b \tan (c+d x))^{3/2}}+\frac {2 \left (2 a^2 A b-4 A b^3+a^3 B+7 a b^2 B\right ) \sqrt {\tan (c+d x)}}{3 b \left (a^2+b^2\right )^2 d \sqrt {a+b \tan (c+d x)}}\\ \end {align*}
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Mathematica [A]
time = 2.04, size = 308, normalized size = 1.26 \begin {gather*} \frac {-\frac {3 B \sqrt {\tan (c+d x)}}{(a+b \tan (c+d x))^{3/2}}+\frac {\left (2 a A b+a^2 B+3 b^2 B\right ) \sqrt {\tan (c+d x)}}{\left (a^2+b^2\right ) (a+b \tan (c+d x))^{3/2}}+\frac {3 \sqrt [4]{-1} b \left (\frac {(a+i b)^2 (i A+B) \text {ArcTan}\left (\frac {\sqrt [4]{-1} \sqrt {-a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{\sqrt {-a+i b}}+\frac {i (a-i b)^2 (A+i B) \text {ArcTan}\left (\frac {\sqrt [4]{-1} \sqrt {a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{\sqrt {a+i b}}\right )+\frac {2 \left (2 a^2 A b-4 A b^3+a^3 B+7 a b^2 B\right ) \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}}{\left (a^2+b^2\right )^2}}{3 b d} \end {gather*}
Antiderivative was successfully verified.
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Maple [B] result has leaf size over 500,000. Avoiding possible recursion issues.
time = 1.48, size = 2976654, normalized size = 12199.40 \[\text {output too large to display}\]
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (A + B \tan {\left (c + d x \right )}\right ) \tan ^{\frac {3}{2}}{\left (c + d x \right )}}{\left (a + b \tan {\left (c + d x \right )}\right )^{\frac {5}{2}}}\, dx \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {{\mathrm {tan}\left (c+d\,x\right )}^{3/2}\,\left (A+B\,\mathrm {tan}\left (c+d\,x\right )\right )}{{\left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )}^{5/2}} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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