Optimal. Leaf size=215 \[ -\frac {2 (B+i A) \sqrt {\tan (c+d x)} (1+i \tan (c+d x))^{-n} (a+i a \tan (c+d x))^n F_1\left (\frac {1}{2};1-n,1;\frac {3}{2};-i \tan (c+d x),i \tan (c+d x)\right )}{d}+\frac {2 (2 B n+i A (2 n+1)) \sqrt {\tan (c+d x)} (1+i \tan (c+d x))^{-n} (a+i a \tan (c+d x))^n \, _2F_1\left (\frac {1}{2},1-n;\frac {3}{2};-i \tan (c+d x)\right )}{d (2 n+1)}+\frac {2 B \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^n}{d (2 n+1)} \]
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Rubi [A] time = 0.49, antiderivative size = 215, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 9, integrand size = 36, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {3597, 3601, 3564, 130, 430, 429, 3599, 66, 64} \[ -\frac {2 (B+i A) \sqrt {\tan (c+d x)} (1+i \tan (c+d x))^{-n} (a+i a \tan (c+d x))^n F_1\left (\frac {1}{2};1-n,1;\frac {3}{2};-i \tan (c+d x),i \tan (c+d x)\right )}{d}+\frac {2 (2 B n+i A (2 n+1)) \sqrt {\tan (c+d x)} (1+i \tan (c+d x))^{-n} (a+i a \tan (c+d x))^n \, _2F_1\left (\frac {1}{2},1-n;\frac {3}{2};-i \tan (c+d x)\right )}{d (2 n+1)}+\frac {2 B \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^n}{d (2 n+1)} \]
Antiderivative was successfully verified.
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Rule 64
Rule 66
Rule 130
Rule 429
Rule 430
Rule 3564
Rule 3597
Rule 3599
Rule 3601
Rubi steps
\begin {align*} \int \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^n (A+B \tan (c+d x)) \, dx &=\frac {2 B \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^n}{d (1+2 n)}+\frac {2 \int \frac {(a+i a \tan (c+d x))^n \left (-\frac {a B}{2}+\frac {1}{2} a (A+2 A n-2 i B n) \tan (c+d x)\right )}{\sqrt {\tan (c+d x)}} \, dx}{a (1+2 n)}\\ &=\frac {2 B \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^n}{d (1+2 n)}+(-i A-B) \int \frac {(a+i a \tan (c+d x))^n}{\sqrt {\tan (c+d x)}} \, dx+\frac {(2 B n+i A (1+2 n)) \int \frac {(a-i a \tan (c+d x)) (a+i a \tan (c+d x))^n}{\sqrt {\tan (c+d x)}} \, dx}{a (1+2 n)}\\ &=\frac {2 B \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^n}{d (1+2 n)}+\frac {\left (a^2 (A-i B)\right ) \operatorname {Subst}\left (\int \frac {(a+x)^{-1+n}}{\sqrt {-\frac {i x}{a}} \left (-a^2+a x\right )} \, dx,x,i a \tan (c+d x)\right )}{d}+\frac {(a (2 B n+i A (1+2 n))) \operatorname {Subst}\left (\int \frac {(a+i a x)^{-1+n}}{\sqrt {x}} \, dx,x,\tan (c+d x)\right )}{d (1+2 n)}\\ &=\frac {2 B \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^n}{d (1+2 n)}+\frac {\left (2 a^3 (i A+B)\right ) \operatorname {Subst}\left (\int \frac {\left (a+i a x^2\right )^{-1+n}}{-a^2+i a^2 x^2} \, dx,x,\sqrt {\tan (c+d x)}\right )}{d}+\frac {\left ((2 B n+i A (1+2 n)) (1+i \tan (c+d x))^{-n} (a+i a \tan (c+d x))^n\right ) \operatorname {Subst}\left (\int \frac {(1+i x)^{-1+n}}{\sqrt {x}} \, dx,x,\tan (c+d x)\right )}{d (1+2 n)}\\ &=\frac {2 B \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^n}{d (1+2 n)}+\frac {2 (2 B n+i A (1+2 n)) \, _2F_1\left (\frac {1}{2},1-n;\frac {3}{2};-i \tan (c+d x)\right ) (1+i \tan (c+d x))^{-n} \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^n}{d (1+2 n)}+\frac {\left (2 a^2 (i A+B) (1+i \tan (c+d x))^{-n} (a+i a \tan (c+d x))^n\right ) \operatorname {Subst}\left (\int \frac {\left (1+i x^2\right )^{-1+n}}{-a^2+i a^2 x^2} \, dx,x,\sqrt {\tan (c+d x)}\right )}{d}\\ &=\frac {2 B \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^n}{d (1+2 n)}-\frac {2 (i A+B) F_1\left (\frac {1}{2};1-n,1;\frac {3}{2};-i \tan (c+d x),i \tan (c+d x)\right ) (1+i \tan (c+d x))^{-n} \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^n}{d}+\frac {2 (2 B n+i A (1+2 n)) \, _2F_1\left (\frac {1}{2},1-n;\frac {3}{2};-i \tan (c+d x)\right ) (1+i \tan (c+d x))^{-n} \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^n}{d (1+2 n)}\\ \end {align*}
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Mathematica [F] time = 19.66, size = 0, normalized size = 0.00 \[ \int \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^n (A+B \tan (c+d x)) \, dx \]
Verification is Not applicable to the result.
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fricas [F] time = 0.54, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left ({\left (A - i \, B\right )} e^{\left (2 i \, d x + 2 i \, c\right )} + A + i \, B\right )} \left (\frac {2 \, a e^{\left (2 i \, d x + 2 i \, c\right )}}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}\right )^{n} \sqrt {\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}, x\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (B \tan \left (d x + c\right ) + A\right )} {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{n} \sqrt {\tan \left (d x + c\right )}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [F] time = 1.83, size = 0, normalized size = 0.00 \[ \int \left (\sqrt {\tan }\left (d x +c \right )\right ) \left (a +i a \tan \left (d x +c \right )\right )^{n} \left (A +B \tan \left (d x +c \right )\right )\, dx \]
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 \[ \int {\left (B \tan \left (d x + c\right ) + A\right )} {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{n} \sqrt {\tan \left (d x + c\right )}\,{d x} \]
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 \[ \int \sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\left (A+B\,\mathrm {tan}\left (c+d\,x\right )\right )\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^n \,d x \]
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 \[ \int \left (i a \left (\tan {\left (c + d x \right )} - i\right )\right )^{n} \left (A + B \tan {\left (c + d x \right )}\right ) \sqrt {\tan {\left (c + d x \right )}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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