Optimal. Leaf size=224 \[ \frac {2 \left (5 a^2-b^2\right ) \sqrt {a+b \tan (c+d x)}}{5 a d \sqrt {\tan (c+d x)}}-\frac {i (-b+i a)^{3/2} \tan ^{-1}\left (\frac {\sqrt {-b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}-\frac {4 b \sqrt {a+b \tan (c+d x)}}{5 d \tan ^{\frac {3}{2}}(c+d x)}-\frac {2 a \sqrt {a+b \tan (c+d x)}}{5 d \tan ^{\frac {5}{2}}(c+d x)}+\frac {i (b+i a)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d} \]
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Rubi [A] time = 0.87, antiderivative size = 224, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 7, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.280, Rules used = {3567, 3649, 3616, 3615, 93, 203, 206} \[ \frac {2 \left (5 a^2-b^2\right ) \sqrt {a+b \tan (c+d x)}}{5 a d \sqrt {\tan (c+d x)}}-\frac {i (-b+i a)^{3/2} \tan ^{-1}\left (\frac {\sqrt {-b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}-\frac {4 b \sqrt {a+b \tan (c+d x)}}{5 d \tan ^{\frac {3}{2}}(c+d x)}-\frac {2 a \sqrt {a+b \tan (c+d x)}}{5 d \tan ^{\frac {5}{2}}(c+d x)}+\frac {i (b+i a)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d} \]
Antiderivative was successfully verified.
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Rule 93
Rule 203
Rule 206
Rule 3567
Rule 3615
Rule 3616
Rule 3649
Rubi steps
\begin {align*} \int \frac {(a+b \tan (c+d x))^{3/2}}{\tan ^{\frac {7}{2}}(c+d x)} \, dx &=-\frac {2 a \sqrt {a+b \tan (c+d x)}}{5 d \tan ^{\frac {5}{2}}(c+d x)}-\frac {2}{5} \int \frac {-3 a b+\frac {5}{2} \left (a^2-b^2\right ) \tan (c+d x)+2 a b \tan ^2(c+d x)}{\tan ^{\frac {5}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}} \, dx\\ &=-\frac {2 a \sqrt {a+b \tan (c+d x)}}{5 d \tan ^{\frac {5}{2}}(c+d x)}-\frac {4 b \sqrt {a+b \tan (c+d x)}}{5 d \tan ^{\frac {3}{2}}(c+d x)}+\frac {4 \int \frac {-\frac {3}{4} a \left (5 a^2-b^2\right )-\frac {15}{2} a^2 b \tan (c+d x)-3 a b^2 \tan ^2(c+d x)}{\tan ^{\frac {3}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}} \, dx}{15 a}\\ &=-\frac {2 a \sqrt {a+b \tan (c+d x)}}{5 d \tan ^{\frac {5}{2}}(c+d x)}-\frac {4 b \sqrt {a+b \tan (c+d x)}}{5 d \tan ^{\frac {3}{2}}(c+d x)}+\frac {2 \left (5 a^2-b^2\right ) \sqrt {a+b \tan (c+d x)}}{5 a d \sqrt {\tan (c+d x)}}-\frac {8 \int \frac {\frac {15 a^3 b}{4}-\frac {15}{8} a^2 \left (a^2-b^2\right ) \tan (c+d x)}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}} \, dx}{15 a^2}\\ &=-\frac {2 a \sqrt {a+b \tan (c+d x)}}{5 d \tan ^{\frac {5}{2}}(c+d x)}-\frac {4 b \sqrt {a+b \tan (c+d x)}}{5 d \tan ^{\frac {3}{2}}(c+d x)}+\frac {2 \left (5 a^2-b^2\right ) \sqrt {a+b \tan (c+d x)}}{5 a d \sqrt {\tan (c+d x)}}-\frac {1}{2} \left (i (a-i b)^2\right ) \int \frac {1+i \tan (c+d x)}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}} \, dx+\frac {1}{2} \left (i (a+i b)^2\right ) \int \frac {1-i \tan (c+d x)}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}} \, dx\\ &=-\frac {2 a \sqrt {a+b \tan (c+d x)}}{5 d \tan ^{\frac {5}{2}}(c+d x)}-\frac {4 b \sqrt {a+b \tan (c+d x)}}{5 d \tan ^{\frac {3}{2}}(c+d x)}+\frac {2 \left (5 a^2-b^2\right ) \sqrt {a+b \tan (c+d x)}}{5 a d \sqrt {\tan (c+d x)}}-\frac {\left (i (a-i b)^2\right ) \operatorname {Subst}\left (\int \frac {1}{(1-i x) \sqrt {x} \sqrt {a+b x}} \, dx,x,\tan (c+d x)\right )}{2 d}+\frac {\left (i (a+i b)^2\right ) \operatorname {Subst}\left (\int \frac {1}{(1+i x) \sqrt {x} \sqrt {a+b x}} \, dx,x,\tan (c+d x)\right )}{2 d}\\ &=-\frac {2 a \sqrt {a+b \tan (c+d x)}}{5 d \tan ^{\frac {5}{2}}(c+d x)}-\frac {4 b \sqrt {a+b \tan (c+d x)}}{5 d \tan ^{\frac {3}{2}}(c+d x)}+\frac {2 \left (5 a^2-b^2\right ) \sqrt {a+b \tan (c+d x)}}{5 a d \sqrt {\tan (c+d x)}}-\frac {\left (i (a-i b)^2\right ) \operatorname {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 )}{d}+\frac {\left (i (a+i b)^2\right ) \operatorname {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 )}{d}\\ &=-\frac {i (i a-b)^{3/2} \tan ^{-1}\left (\frac {\sqrt {i a-b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}+\frac {i (i a+b)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {i a+b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}-\frac {2 a \sqrt {a+b \tan (c+d x)}}{5 d \tan ^{\frac {5}{2}}(c+d x)}-\frac {4 b \sqrt {a+b \tan (c+d x)}}{5 d \tan ^{\frac {3}{2}}(c+d x)}+\frac {2 \left (5 a^2-b^2\right ) \sqrt {a+b \tan (c+d x)}}{5 a d \sqrt {\tan (c+d x)}}\\ \end {align*}
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Mathematica [A] time = 1.67, size = 197, normalized size = 0.88 \[ \frac {\frac {2 \sqrt {a+b \tan (c+d x)} \left (\left (5 a^2-b^2\right ) \tan ^2(c+d x)-a^2-2 a b \tan (c+d x)\right )}{a \tan ^{\frac {5}{2}}(c+d x)}-5 \sqrt [4]{-1} (-a+i b)^{3/2} \tan ^{-1}\left (\frac {\sqrt [4]{-1} \sqrt {-a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )+5 \sqrt [4]{-1} (a+i b)^{3/2} \tan ^{-1}\left (\frac {\sqrt [4]{-1} \sqrt {a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{5 d} \]
Antiderivative was successfully verified.
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fricas [F(-1)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [F(-1)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 0.80, size = 1346038, normalized size = 6009.10 \[ \text {output too large to display} \]
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 \[ \int \frac {{\left (b \tan \left (d x + c\right ) + a\right )}^{\frac {3}{2}}}{\tan \left (d x + c\right )^{\frac {7}{2}}}\,{d x} \]
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 \[ \int \frac {{\left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )}^{3/2}}{{\mathrm {tan}\left (c+d\,x\right )}^{7/2}} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F(-1)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]
Verification of antiderivative is not currently implemented for this CAS.
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